What is the integral of 22sec^4(θ)tan^2(θ) ?

The integral of 22sec^4(θ)tan^2(θ) is 22((tan^3(θ))/3+(tan^5(θ))/5)+C

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integral of 22sec^4(θ)tan^2(θ)

\int 22 sec^{4} (θ) tan^{2} (θ) d θ

22 (\frac{tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5}) + C

Solution steps

\int 22 sec^{4} (θ) tan^{2} (θ) d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 22 \cdot \int sec^{4} (θ) tan^{2} (θ) d θ

Rewrite using trig identities

= 22 \cdot \int (1 + tan^{2} (θ)) sec^{2} (θ) tan^{2} (θ) d θ

Apply u-substitution

= 22 \cdot \int u^{2} (1 + u^{2}) d u

Expand

u^{2} (1 + u^{2}) : u^{2} + u^{4}

= 22 \cdot \int u^{2} + u^{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 22 (\int u^{2} d u + \int u^{4} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int u^{4} d u = \frac{u ^{5}}{5}

= 22 (\frac{u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = tan (θ)

= 22 (\frac{tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5})

Add a constant to the solution

= 22 (\frac{tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5}) + C

s l o p e (4.1) (1.1)

\frac{d}{d x} (3 tan (π - x))

\frac{d}{d x} (\frac{x ^{3} e ^{x}}{x})

\frac{d}{d y} (\frac{x ln ( \frac{x y}{z} )}{z})

d er i v a t i v e 12 x - 2 x

What is the integral of 22sec^4(θ)tan^2(θ) ?
The integral of 22sec^4(θ)tan^2(θ) is 22((tan^3(θ))/3+(tan^5(θ))/5)+C